Show that $$\int\limits_0^\infty\frac{1}{t}(\cos(at)-\cos(bt))dt=\ln(b/a),\,a,b>0.$$

Thanks to wikipedia I know that

$$\int\limits_0^\infty\frac{1}{t}(\cos(at)-\cos(bt))\,dt \overset{?}{=}\int\limits_{0}^\infty\left[\frac{p}{p^2+a^2}-\frac{p}{p^2+b^2}\right]=\left.\frac{1}{2}\ln\left[\frac{p^2+a^2}{p^2+b^2}\right]\right|_{0}^\infty=\ln b-\ln a.$$ I am having a hard time understanding the jump between the equality with the question mark above it.

  • $\begingroup$ The Laplace Transform of the cosine function is $\int_0^\infty \cos(at)e^{-pt}\,dt=\frac{p}{p^2+a^2}$. $\endgroup$ – Mark Viola Feb 3 '16 at 4:46


$$\frac1{t} = \int_0^{\infty} dp \, e^{-p t}$$

Then the integral is

$$\begin{align}\int_0^{\infty} dt \, \frac{\cos{a t}-\cos{b t}}{t} &= \int_0^{\infty} dt \, (\cos{a t} - \cos{b t})\int_0^{\infty} dp \, e^{-p t} \\ &= \int_0^{\infty} dp \, \left (\int_0^{\infty} dt \, \cos{a t} \, e^{- p t} - \int_0^{\infty} dt \, \cos{b t} \, e^{- p t}\right ) \\ &= \int_0^{\infty} dp \, \left (\frac{p}{p^2+a^2} - \frac{p}{p^2+b^2} \right ) \end{align} $$

It seems you got the rest. Note that, in the second line, we can reverse the order of integration because the integrals involved are convergent. Also note that, in the third line, as @DrMV points out, the expressions are Laplace transforms of the respective cosine functions.

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  • $\begingroup$ To be honest, I don't understand what you mean by "the integrals involved are convergent". For Fubini, we would need integrability of $\int |\cos(at) - \cos(bt)| e^{-pt} \, dt \, dp<\infty$. I don't think that (don't see why) this is the case. $\endgroup$ – PhoemueX Feb 3 '16 at 9:50

Have you heard about Frullani's Integrals? Otherwise you can see here for different techniques.

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  • 1
    $\begingroup$ This reference Frullani relaxes the condition on $f$. $\endgroup$ – Mark Viola Feb 3 '16 at 5:08

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